# CS173: Intro to Computer Science - Boolean Expressions

# Activity Goals

The goals of this activity are:- To be able to write a boolean expression using variables of various types

# Supplemental Reading

Feel free to visit these resources for supplemental background reading material.# The Activity

## Directions

Consider the activity models and answer the questions provided. First reflect on these questions on your own briefly, before discussing and comparing your thoughts with your group. Appoint one member of your group to discuss your findings with the class, and the rest of the group should help that member prepare their response. Answer each question individually from the activity on the Class Activity Questions discussion board. After class, think about the questions in the reflective prompt and respond to those individually in your notebook. Report out on areas of disagreement or items for which you and your group identified alternative approaches. Write down and report out questions you encountered along the way for group discussion.## Model 1: Representing Truth as a Variable to Make Decisions in Programs

**Type**

**Purpose**

**Example**

`&&`

AND two booleans (compute true if both are true)

`boolean b = false;`

`boolean x = true && b; // false`

`||`

OR two booleans (compute true if at least one is true)

`boolean x = (true || false); // true`

`!`

NEGATE a boolean

`boolean x = !(true || false); // = !(true) = false`

`==`

Check if two values are equal

`boolean x = (10 == 5); // false`

`!=`

Check if two values are not equal. This is the same as

`!(x == y)`

`boolean x = (10 != 5); // true`

### Questions

- What is the result of
`(a || b) && (c || d)`

if`a = true`

,`b = true`

,`c = false`

,`d = false`

? - If
`a`

is true in the example above, is it necessary to evaluate`b`

at all? - DeMorganâ€™s Law allows you to simplify a boolean expression by "factoring out" a negation, and flipping an AND to an OR (and vice-versa). For example,
`(!a && !b)`

is equivalent to`!(a || b)`

. The reverse procedure also works - negating the outside, negating each term on the inside, and flipping the operator:`!(a || b)`

is equivalent to`(!a && !b)`

. Re-write`!(a && !b)`

using DeMorganâ€™s Law. - Write a program to compute
`(!a && !b)`

for two`boolean`

variables, and then to compute the DeMorgan's Law version of that expression. Try it with all four combinations of boolean variables (true/true, false/false, true/false, false/true), and print both results to verify that they are equivalent. - In the Venn diagram above, assign
`a`

to the left circle, and`b`

to the right circle. Fill in`!a || !b`

. Now fill in`!(a && b)`

. How do they differ?